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In the figure above, the two square regions have areas 16 and 25, resp

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In the figure above, the two square regions have areas 16 and 25, resp  [#permalink]

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New post 31 Oct 2018, 18:19
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In the figure above, the two square regions have areas 16 and 25, respectively. What is the area of the shaded triangular region?

A. 6
B. 8
C. 9
D. 12
E. 15


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Re: In the figure above, the two square regions have areas 16 and 25, resp  [#permalink]

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New post 31 Oct 2018, 19:41
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Area of the first square = 16. Side = 4.
Area of the tilted square = 25. Side = 5.
The side of the tilted square becomes the hypotenuse of the triangle = 5.
The side of the first square becomes the height of the triangle. = 4.
Then by Pythagoras theorem, the base is 3.
Area of the triangle = 1/2*4*3 = 6.

A is the answer.
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Re: In the figure above, the two square regions have areas 16 and 25, resp  [#permalink]

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New post 05 Nov 2019, 12:52
Hi All,

We're told that the two squares have areas 16 and 25, respectively. We're asked for the AREA of the shaded triangular region. This is an example of how the GMAT writers sometimes like to 'hide' the special Right Triangles (re: 3/4/5, 5/12/13, 30/60/90 and 45/45/90) in Geometry questions - and if you recognize the patterns involved, then you can answer this question relatively quickly without too much work.

To start, based on the areas of the two squares, we know that their respective sides are 4s and 5s. Thus, the right triangle has a height of 4 and a hypotenuse of 5. You probably recognize this as a 3/4/5 right triangle, but even if you didn't, you can use the Pythagorean Theorem to prove it:

A^2 + B^2 = C^2
A^2 + 4^2 = 5^2
A^2 + 16 = 25
A^2 = 9
A = 3... since this is a shape, it can't have a "negative" side (so -3 is not an option).

With a height of 4 and a base of 3, we can determine the AREA of the shape:
A = (1/2)(base)(height)
A = (1/2)(3)(4)
A = 6

Final Answer:

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Re: In the figure above, the two square regions have areas 16 and 25, resp   [#permalink] 05 Nov 2019, 12:52
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